Докажите, что: 1) 2) 3)

Решите задачу:

$\lim_{n \to +\infty} \frac{2x+5}{3x}= \lim_{n \to \infty} \frac{\frac{2x}x+\frac{5}x}{\frac{3x}x}= \lim_{n \to \infty} \frac{2+\frac{5}x}{3}=\frac{2}3\\\\ \lim_{n \to +\infty} \frac{3x^2+6}{x^2}=\lim_{n \to +\infty} \frac{\frac{3x^2}{x^2}+\frac{6}{x^2}}{\frac{x^2}{x^2}}=\lim_{n \to +\infty} \frac{3+\frac{6}{x^2}}{1}=3\\\\ \lim_{n \to +\infty} \frac{5x+14}{x+2}=\lim_{n \to +\infty} \frac{\frac{5x}x+\frac{14}x}{\frac{x}x+\frac{x}2}=\lim_{n \to +\infty} \frac{5+\frac{14}x}{1+\frac{2}x}=5$